Codeforces 474D Flowers

We saw the little game Marmot made for Mole's lunch. Now it's Marmot's dinner time and, as we all know, Marmot eats flowers. At every dinner he eats some red and white flowers. Therefore a dinner can be represented as a sequence of several flowers, some of them white and some of them red.

But, for a dinner to be tasty, there is a rule: Marmot wants to eat white flowers only in groups of size k.

Now Marmot wonders in how many ways he can eat between a and b flowers. As the number of ways could be very large, print it modulo1000000007 (109 + 7).

Input

Input contains several test cases.

The first line contains two integers t and k (1 ≤ t, k ≤ 105), where t represents the number of test cases.

The next t lines contain two integers ai and bi (1 ≤ ai ≤ bi ≤ 105), describing the i-th test.

Output

Print t lines to the standard output. The i-th line should contain the number of ways in which Marmot can eat between ai and bi flowers at dinner modulo 1000000007 (109 + 7).

Sample test(s)

input

3 2
1 3
2 3
4 4

output

6
5
5

Note

• For K = 2 and length 1 Marmot can eat (R).
• For K = 2 and length 2 Marmot can eat (RR) and (WW).
• For K = 2 and length 3 Marmot can eat (RRR), (RWW) and (WWR).
• For K = 2 and length 4 Marmot can eat, for example, (WWWW) or (RWWR), but for example he can't eat (WWWR).

解题思路：简单dp。 dp[i][0]:表示第i朵是红花的方案数,dp[i][1]表示第i朵是白花的方案数,具体的状态转移方程见代码.

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int mod = 1000000000 + 7;
const int maxn = 100010;
int dp[maxn][2];
int sum[maxn];
int n, k;

void init() {
    dp[0][0] = 1;
    dp[0][1] = 0;
    sum[0] = 0;
    for(int i = 1; i <= 100000; ++i) {
        dp[i][0] = (dp[i-1][0] + dp[i-1][1]) % mod;
        if(i >= k) {
            dp[i][1] = (dp[i-k][0] + dp[i-k][1]) % mod;
        } else {
            dp[i][1] = 0;
        }
        sum[i] = (sum[i-1] + dp[i][0]) % mod;
        sum[i] = (sum[i] + dp[i][1]) % mod;
    }
    return ;
}


int main() {

    int a, b;
    scanf("%d %d", &n, &k);
    init();
    for(int i = 1; i <= n; ++i) {
        scanf("%d %d", &a, &b);
        if(a > b) {
            swap(a, b);
        }
        printf("%d\n", ((sum[b]-sum[a-1])%mod+mod)%mod);
    }
    return 0;
}